Energy consumption of a cold storage was measured for different storage temperatures. Suction temperature and pressure temperature of the compressor and working time of the compressor were determined to reach evaporator setup temperatures. An axial fan located back of the evaporator was used to distribute the cooled air into the cold store. An electrical heater was used to defrost. The compressor suction temperature of ammonia vapour variedbetween273K–271Kand 305K–308K respectively. Compressor suction pressure(p1)=3.5 Kg/cm 2 and discharge pressure (p2)=10.5Kg/cm2

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Energy Wastage Estimation of a Cold Storage
Dr. N. Mukhopadhyay 1
, Debasish Mazumdar ₂
, Subir Biswas₃
1
Assistant professor, Department of Mechanical Engineering, Jalpaiguri Govt.Engg. College,
2
M.Tech. Scholar, Department of Mechanical Engineering, Jalpaiguri Govt.Engg. College,
₃
B.Tech ,Final year student of Mechanical Engineering, Jalpaiguri Govt.Engg. College
ABSTRACT:
Energy consumption of a cold storage was measured for different storage temperatures. Suction temperature and
pressure temperature of the compressor and working time of the compressor were determined to reach
evaporator setup temperatures. An axial fan located back of the evaporator was used to distribute the cooled air
into the cold store. An electrical heater was used to defrost. The compressor suction temperature of ammonia
vapour variedbetween273K–271Kand 305K–308K respectively. Compressor suction pressure(p1)=3.5 Kg/cm2
and discharge pressure (p2)=10.5Kg/cm2
KEY WORDS: Cold storage, major components of a cold storage, energy consumption, energy wastage,
programming.
I. INTRODUCTION
Energy auditing in a integral part of energy conservation and energy management is also part and parallel of
conservation. Damage and supply gap is large energy to lead to similar natural defects, Energy disaster such as
Tsunami and earth quake. The next generation generating yet to come will be completely light blind. It is
because power never be available after this disaster and not ever rehabilitate the reconstruction of buildings. To
avoid the energy calamity proposed auditing report use the innovative energy utilization schemes through which
the ferocious of situation might blindness can be eradicated. Cold Storage is a special kind of room, the
temperature of, which is kept very low with the help of machines and precision instruments. India is having a
unique geographical position and a wide range of soil thus producing variety of fruits and vegetables like apples,
grapes, oranges, potatoes, chillies, ginger, etc. Marine products are also being produced in large quantities due
to large coastal areas. The present production level of fruits and vegetables is more than 100 million MT and
keeping in view the growth rate of population and demand, the production of perishable commodities is
increasing every year. The cold storage facilities are the prime infrastructural component for such perishable
commodities. Besides the role of stabilizing market prices and evenly distributing both on demand basis and
time basis, the cold storage industry renders other advantages and benefits to both the farmers and the
consumers. The farmers get opportunity of producing cash crops to get remunerative prices. The consumers get
the supply of perishable commodities with lower fluctuation of prices. Commercially apples, potatoes, oranges
are stored on large scale in the cold storages. Other important costly raw materials like dry fruits, chemicals,
essences and processed foods like fruit juice/pulp, concentrate dairy products, frozen meat, fish and eggs are
being stored in cold storages to regulate marketing channels of these products. Energy consumption of an
experimental cold storage was measured for different storage temperatures. Suction temperature and pressure
temperature of the compressor and working time of the compressor were determined to reach evaporator set up
temperatures.
II. MODEL DEVELOPMENT:
The salient components of vapour compression refrigeration system used in Indian cold storage
system are: evaporator, compressor, condenser and manually operated expansion valve. The type of the cold
storage are
1) Vapour campression refrigeration cycle is used
2)Ammonia refrigerent is used in the cycle
For the analysis of the various parameter of the Jalpaiguri cold storage we make some assumption
1. The compression and expansion process are an isentropic process ie. Entropy of those process are
remains same
2. The ammonia vapour behaves as an ideal gas
RESEARCH ARTICLE OPEN ACCESS

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3. Specific heat, specific heat ratio and local gas constant does not change in different temperature and
pressure.
4. The specific heat capacities at constant pressure and constant volume processes, and the ratio of
specific heat and the individual gas constant - R – for ammonia vapour at 20 °c and 1a.t.p are given
below
C(p)= 2.19; C(v)= 1.66 ; ɣ= 1.31 ; R= 0.53
A typical schematic diagram of the refrigeration stem is shown below
Cold storage
From the above assumption we can calculate following parameters
Compression work Wc=n/𝑛 − 1(P₂V₂-P₁V₁)
=n/n – 1(RT₂ - RT₁)
= 78.38 KJ/ Kg of ammonia
Expansion Work Wₑ= n/n- 1(RT₃ - RT₄)
= 76 KJ / Kg of ammonia
net work done W = W(c) – W(e)
= (78.38 – 76) KJ / Kg of ammonia
= 2.38 KJ / Kg of ammonia
Heat absorbed from the evaporator per kg of ammonia
Qₑ = mCp(T₁ - T₄)
=4.38 KJ / Kg of ammonia
C.O.P = Heat absorbed / Work done
= 4.38 / 2.38
= 1.84
Volume of ammonia per cycle of compression per cylinder
V₁=
π
4
𝐷²L [from the observation table we can find bore and stroke]
= 0.002796 m³ ≈ 0.0028m³
So work of compression per cycle per cylinder
W̕c =
𝑛
𝑛−1
(P₂V₂ - P₁V₁)

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=
𝑛
𝑛−1
P₁V₁{(
𝑃₂
𝑃₁
)
𝑛
𝑛−1 − 1}
= 4.226×320×0.0028×(1.3846 - 1)
= 1.456KJ / cycle per cylinder
≈1.5 KJ / cycle per cylinder
There are two cylinder of a KC 4 type compressor and average R.P.M = 1400
SoWork of compressor per second
W̕c = (1.5 × 2 × 1400)/ 60 KJ / sec
= 70 KJ/ sec
Mass flow rate of ammonia per second
=
𝐖̕ 𝐜
Wc
=
70
78.38
= 0.893 Kg / sec
Ton of refrigeration
= mCp(T₁ - T₂)/3.5
= 1.178TR
≈ 1.2 TR
Heat dissipated from the condenser
Qc = mCp(T₂- T₃)
= 5.867 KJ / sec
≈ 6 KW
OBSERVATION:
 Suction pressure of the compressor P₁= 319.9 KPa≈320KPa
 Suction temperature of compressor T₁=273K
 Discharge pressure of the compressor P₂=1226.25KPa
 Discharge temperature of compressor T₂=308K
 Discharge pressure of the condenser P₃=P₂=1226.25KPa
 Discharge temperature of the condenser T₃=305K
 DATA COLLECTION
 Major equipments considered to calculate the energy waste estimation
Equipment name Number Stand by
Compressor motor 4p 2
Condenser motor 3p 1
Pump 3p 1
 Minor equipments considered to calculate the energy waste estimation
No of fans Place Power
320p Inside the chamber 72w
36p On the office 85w
50p Outside of chamber 72w
No of light Power Place
430p 60w Inside the chamber
50p 40w In office
10p 280w Hallozen
MOTOR SPECIFICATION
Motor specification= 3 phase induction motor
Company name= crompton graves
m/c no= NADJ54-2RS
RPM- 1460
V=415 +(-) 10% A=27A
BPE=90% HZ=50+(-) 5%
FRAME ND160L

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PUMP SPECIFICATION
PUMP SPECIFICATIONS
COMPANY= KSB PUMPS LTD
SL NO= 9850877
Q = 500m3/h H=52M
N-2980RPM
IN0352D-A52*74
COMPRESSOR SPECIFICATION
KIRLOSKAR PNEUMATIC CO.LTD
MODEL – K.C-4 SL.NO-338K1461
BORE =160mm STROKE=110mm
MAX PR=21KG/C
RPM=1460 SPEC NO= 338001
Compressor motor running time calculation
Compressor motor 1 Compressor motor 2
Date Start
time
Stop
time
Next
day
stop
time
Running
time
Date Start
time
Stop
time
Next
day
stop
time
Running
time
23-
10-15
10.25AM 8.50PM 10H25M
11.10PM 7.30AM 8H20M
24- 11AM 8.50PM 9H50M
11PM 9.10AM 10H10M
25- 2.10PM 5.40PM 3H30M 25-2 10.20AM 7.10PM 8H50m
11.10PM 6.20AM 7H10M
26- 10.50AM 3.40PM 4H50M 26-2 3.20AM 4.30PM 13H10M
11.10PM 5.10AM 6H 6.20PM 2.30AM 8H10M
27- 10.20AM 5.10PM 6H50M 27-2 7.25AM 4.05PM 8H40M
11.10PM 6.20AM 7H10M 8PM 2.10AM 6H10M
28- 10.50AM 3.40PM 4H50M 28-2 5AM 1.20PM 8H20M
11PM 8.10AM 9H10M 4.30PM 2.50AM 10H20M
29 10AM 2.10PM 4H10M 1-3 6.10AM 10AM 3H50M
11.10PM 3.10AM 4H 4.20PM 8.20PM 4H
30 6.40AM 10.50AM 4H10M 2-3 10.AM 2PM 4H
11.10PM 2.30AM 3H20M 4.50PM 9.40PM 4H50M
31 10AM 2PM 4H 3-3 6.10AM 10.10AM 4H
11PM 3AM 4H 4PM 8PM 4H
1-11 9.20AM 3.50PM 5H30M 4-3 5.40AM 11AM 5H20M
2.10PM 7.30PM 5H10M 11.10PM 5.25AM 6H15M
2 6.10AM 9.50AM 3H40M 5-3 10.10AM 2.10PM 4H
4.40PM 9.20PM 4H40M 11.10PM 3.40AM 4H30M
3 6.20AM 11.30AM 4H10M 6-3 12.40PM 5PM 4H20M
8.10PM 11.30PM 3H20M
4 10.10AM 2.30PM 4H20M 7-3 3AM 7.10AM 4H10M
11.10PM 8AM 9H10M
5 10.10AM 6.40PM 8H30M
11.10PM 6.10AM 7H
6 9-3 10.20AM 4.40PM 6H20M
11.10PM 4.20AM 5H10M
7 6.25AM 10.30AM 4H05M 10-3 12.20PM 4.50PM 4H30M
4.30PM 9.50PM 5H20M 11.10PM 3.20AM 4H10M
8 10AM 2.30PM 4H30M 11-3 6AM 10.50AM 4H50M
11.10PM 3.40AM 4H30M 4PM 8.20PM 4H20M

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9 10.10AM 3.30PM 5H20M 12-3 6.10AM 11.50AM 5H30M
11.10PM 3.50AM 4H40M 4.10PM 9PM 4H50M
10 10AM 2.30PM 4H30M 13-3 6AM 10.40AM 4H40M
11PM 3.40AM 4H40M 11PM 3.20AM 4H20M
11 10.10AM 2.50PM 4H40M 14-3 6.10AM 10.20AM 4H10M
11.10PM 4.50AM 5H40M 4.20PM 9.40PM 5H20M
12 10AM 2.30PM 4H30M 15-3 6AM 10.20AM 4H20M
11PM 3.40AM 4H40M 4PM 8.20PM 4H20M
13 10.10AM 2.20PM 4H10M 16-3 6.20AM 10.50AM 4H30M
11PM 3.30AM 4H30M 11.10PM 3.30AM 4H20M
14 10AM 2.30PM 4H30M 17-3 6.10AM 10.20AM 4H10M
11.10PM 3.50AM 4H40M 11.10PM 3.30AM 4H20M
15 10.30AM 4.10PM 5H40M 18-3 6AM 11.30AM 5H30M
11.20AM 3.30AM 4H10M 4PM 8.20PM 4H20M
16 10AM 2.20PM 4H20M 19-3 6.10AM 10.40PM 4H30M
11PM 3.30AM 4H30M 11.10PM 3.50AM 4H40M
17 10.20AM 2.40PM 4H20M 20-3 6.30AM 11.10AM 4H40M
11.10PM 3.40AM 4H30M 4.10PM 8.20PM 4H10M
18 10AM 2.30PM 4H30M 21-3 6.10AM 10.30AM 4H20M
11.10PM 4.50AM 5H40M 4.20PM 8.50PM 4H30M
19 10.20AM 2.40PM 4H20M 22-3 6AM 10.20AM 4H20M
11.20PM 3.50AM 4H30M 4.10PM 8.50PM 4H30M
20 10AM 2.20PM 4H20M 23-3 6.20AM 10.50AM 4H30M
11.10PM 3.20AM 4H10M 11.10PM 3.50AM 4H40M
21 6.10AM 11.50AM 5H40M
11.10PM 4.30AM 5H20M
22 6.05AM 12.30PM 6H25M
11.10PM 5.30AM 6H20M
23 10.20AM 2.40PM 4H20M
11.10PM 420AM 5H10M
TOTAL RUNNING TIME - 603.6H; APPROXIMATLY WE CAN TAKE 604 H
CALCULATION
Power consumed by the light and fan can be calculated by the given formula
KWh =
𝑃 𝑤𝑎𝑡𝑡 ×𝑕𝑜𝑢𝑟 ×𝑛𝑜 𝑜𝑓 𝑓𝑎𝑛 𝑜𝑟 𝑙𝑖𝑔𝑕𝑡
1000
POWER CONSUMED BY THE FAN
NO
OF
FANS
POWER
FAN TYPE RUNNING TIME IN
24H
RUNNING
TIME IN ONE
MONTH
KWH IN
24H
KWH IN
ONE
MONTH
320 72 INDUSTRIAL 604 13916.16
50 72 INDUSTRIAL 12 43.2 1339.2
35 85 USHA 7 20.825 645.57
TOTAL POWER CONSUMED BY THE FAN IN ONE MONTH =15900.93
POWER CONSUMED BY THE LIGHT
NO OF LIGHT POWER RUNNING TIME IN 24H KWH IN 24 H KWH IN ONE MONTH
500P 60W 2H 60 1860
50P 40W 4H 8 248
5P 250W 7H 8.75 271.25
TOTAL POWER CONSUMED IN 24 HOURS=76.75
TOTAL POWER CONSUMED IN ONE MONTH =2379.25
Power consumed by the 3 phase induction motor can be calculated by the given formula
P(watt)= V × I× 𝐜𝐨𝐬 ∅

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V= voltage of the motor
I = current flow through the motor
cos ∅ = power factor = 0.85
We can find V & I from the volt meter and ammeter
Power factor of W.B.S.E.D.C.L is 0.85 (we found these from google)
POWER CONSUMED BY THE CONDENSER MOTOR
MOTOR
NAME
VOLTAGE(V) CURRENT
FLOW(AMP)
POWER
FACTOR
watt KWh
C.G 415 12 0.85 4233 4.233
So power consumed by the condenser motor in one hour = 4.233 KWh
POWER CONSUMED BY THE MOTOR IN ONE MONTH = 2556.732 KWh
POWER CONSUMED BY THE COMPRESSOR MOTOR
MOTOR
NAME
VOLTAGE(V) CURRENT
FLOW(AMP)
POWER
FACTOR
watt KWH
1 415 125 0.85 44093.75 44.09
Power consumed by the compressor motor in one hour = 44.09 KWh
POWER CONSUMED BY THE COMPRESSOR MOTOR IN ONE MONTH = 26630.36KWh
We make a program of P(watt)= V × I× cos ∅ from where we can make a graph of I Vs P
PROGRAMMING:
#include<stdio.h>
#include<conio.h>
#include<math.h>
Void main()
{
Float P,I;
Int V=415, n=10 , i;
Float cos ∅ =0.85;
For(i =0; i<n; i++)
{
Printf{“n Enter the the value of I”};
Scanf(“%f”, &I);
P = v*I* cos ∅;
Printf(“I=%f, P=%”, I, P);
}
Getch();
}
RESULT & DISCUSSION:
In the above program we give the input value of I =12 to I= 7
And take the output P and make a graph P Vs I
I(amp) 12 11.5 11 10.5 10 9.5 9 8 7
P(KWh) 4.23 4.05 3.88 3.7 3.52 3.35 3.17 2.8 2.45

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Form
data we
can say
that
OCTOBER 23 to NOVEMBER 23 unit runs more over 24 hours.
Total running time of the plant = 604 hours
From the electricity bill 23rd
October to 23rd
November unit consumed by the plant
= 21515.5KWh(normal) + 2484.25KWh(peak) + 24442.25KWh(off peak)
= 48442KWH
Theoretical energy consumed by the plant in one month
Device Condenser motor Compressor motor Light Fan
Unit consumed in
one month
2556.73 26630.36 2379.25 15900.93
Unit consumed in
one day
82.47 859.04 76.75 512.93
So total theoretical energy consumed by the plant in one month =47467.27 KWH
Energy wastage in one month = (48442-47467.27)KWH
=974.73 KWH
There is a big difference in actual energy consumed and theoretical energy consumed obviously there is a
energy wastage. We must pay extra cost for these energy wastage. Average industrial bill = 7.5 Rs
ENERGY WASTAG AND EXTRA COST REQUIRED HAS BEEN CALCULATED IN THE GIVEN BOX
Energy wastage In one hour In one day In one month In one year
KWh 1.31KWh 31.44KWh 974.73KWh 11696.76KWh
Extra cost required ₨=9.825 ₨= 235.8 ₨=7310.47 ₨=87725.7
III. CONCLUSION
It has been observed that the actual energy consumption of a cold storage 48442 KWh per month and theoretical
energy consumption is 47467.27 KW per month. So, a gap of 974.73 KWh per month is found between the
theoretical and actual energy consumption by a cold storage plant. If this gap is fulfilled then a savings of Rs
7311 /- per month will be obtained which will amount to Rs 87726 /- per year.
To save energy the following practices must be implemented
1. Reducing Heat Loads
2. Uses of latest energy saving equipments
3. Proper Insulation

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ISSN: 2248-9622, Vol. 5, Issue 12, (Part - 3) December 2015, pp.166-173
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4. Efficient maintenance practices
5. Automation & Integration
6.Repless all lights and Uses of LED
There are other factor which contributes to the total energy consumption, which are (i) lights (total 556 pcs) , (ii)
fans (total 405 pcs). To increase the efficiency, conventional lights will replaced by energy efficient devices that
is LED, CFL etc.
If we replace three compressors for three chambers with a single equivalent compressor which will deliver the
same amount of load, then maximumenergy will be saved
REFERENCE :
[1]. Condenser and Evaporator-Version I ME IIT Kharagpur Lesson 22.
[2]. Dr. N.Mukhopadhyay-Theoretical Convective Heat Transfer Model Development of Cold Storage
Using Taguchi Analysis.ISSN:2248-9622,Vol.5,Issue 1,(part-6)January2015.
[3]. M.S. Soylemez, M. Unsal(1997) Optimum Insulation Thickness for Refrigerant applications, Energy
Conservation and Management 40(1999) 13-21.
[4]. ASHRAE Handbook of Fundamentals, 1993.
[5]. B.Kundu- international communications in heat and mass transfer,2002.(An analytica study of the effect
of dehumidification of air on the performance and optimization of straight tapered fins.
[6]. Process heat transfer by DONAL Q. KERN (MCGRAW-HILL).
[7]. Mathematics department of Massachusettes institute of technology.
[8] Refrigeration & Air conditioning by R.S Khurmi